Talk:Fourier series

recent edits introduce numerous errors and inconsistencies

I am restoring basically the 15-Feb version with some improvements. The recent edits warrant discussion and hopefully a consensus. Bob K (talk) 16:07, 20 February 2023 (UTC)

OK, I undid the undo. There are two problems I have with the Feb 15th version. First, it incorrectly defines the Fourier series, instead giving formulas for the partial sum. This is (IMO) a serious issue, because there is a lot to talk about in terms of which partial sums you should take and why, which cannot be discussed if you define the series immediately to be the partial sum. The second issue is that very early on in the article into a discussion of the cross correlation function, which was causing some confusion, so I tried to move it. Finally, the revert undid other peoples edits that had nothing to do with the change I introduced, which are presumably the objectionable ones. I will read through it again and try to fix any errors or inconsistencies I see, but I do think this is a better starting point. Thenub314 (talk) 17:22, 20 February 2023 (UTC)

First of all, I am not committed to the partial sum representation. That is someone else's contribution that I have tried to respectively maintain, because it seems harmless enough (to me). Secondly, I appreciate that you merely "moved" the cross-correlation stuff. And that probably would have been fine, except for the other problems that resulted. Thirdly, yes I realize I undid some other minor edits, but I think they actually do have to do with your changes. Whether or not that is strictly true, they are very minor ones that could easily be redone... not justification for a complete undo.

While you are "reading through it again", consider:

• As you no doubt know, the Fourier series is not limited to periodic s(x). One cycle of it can represent a function in just an interval (or with compact support, as it is sometimes described). Your restriction to periodic s(x) is unnecessary, and inconsistent with Fig. 1.
• The section Complex-valued functions is deleted, and not properly replaced with anything explicit.
• Do you seriously believe that a consensus of editors will prefer the amplitude of component $n,$ in terms of its subcomponents, to be $a_{n}={\sqrt {A_{n}^{2}+B_{n}^{2}}}$ instead of $A_{n}={\sqrt {a_{n}^{2}+b_{n}^{2}}}$ ?

There's more, but you'll presumably catch some of them in your read-through.
--Bob K (talk) 23:52, 20 February 2023 (UTC)

"This is a property that similarly have this property"?

This sentence doesn't make any sense as-is. Unfortunately I don't really understand what it could mean. I am going to remove it for the moment. pglpm (talk) 10:00, 23 February 2023 (UTC)

Oh, I must of clobbered that sentence trying to edit it, thanks Thenub314 (talk) 13:26, 23 February 2023 (UTC)

Error in Hilbert space section?

It seems to me that the Hilbert space being considered must be $L^{2}(-\pi ,\pi )$, since otherwise we need $f(-\pi )=f(\pi )$ by periodicity. jajaperson (talk) 05:25, 6 March 2024 (UTC)

"Hilbert Spaces and Fourier analysis" listed at Redirects for discussion

The redirect Hilbert Spaces and Fourier analysis has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2024 March 28 § Hilbert Spaces and Fourier analysis until a consensus is reached. 1234qwer1234qwer4 01:32, 28 March 2024 (UTC)

Continuously differentiable functions

Back in 2008 in this edit User:Silly rabbit put "We have already mentioned that if ƒ is continuously differentiable, then $n{\hat {f}}(n)$ converges to zero as n goes to infinity. It follows, essentially from the Cauchy-Schwarz inequality, that the Fourier series of ƒ converges absolutely. Since the Fourier series converges in the mean to ƒ, it must converge uniformly:" and then he states a theorem.

I don't understand what is meant. (I tried sending an e-mail but got no answer.) What does this have to do with the Cauchy-Schwarz inequality? How do you use that to show this?

It is certainly possible for a sequence of coefficients to go to zero, and for n times them to go to zero, without the sequence being summable. For example, if $a_{n}=1/(n\log n)$ (for $n>1$) then they go toward zero, as does $na_{n}$, but the sum of the $a_{n}$'s goes to infinity.

Also, what does "converges in the mean" mean?

User:Charles Matthews, maybe you can help?

Eric Kvaalen (talk) 16:10, 7 September 2024 (UTC)

I did reply to the email. The key point is that if $f$ is continuously differentiable, the sequence Fourier coefficients of $f'$, $n{\hat {f}}(n)$, belongs to $\ell ^{2}$. Therefore, ${\hat {f}}(n)$ belongs to $\ell ^{1}$, because (by Cauchy)

$\left(\sum |{\hat {f}}(n)|\right)^{2}\leq \sum {\frac {1}{n^{2}}}\cdot \sum |n{\hat {f}}(n)|^{2}.$

Converges in mean is convergence in $L^{1}$. -- Silly rabbit (talk) 16:47, 7 September 2024 (UTC)

@Silly rabbit: Ah, I found your e-mail. Stupid Hotmail put it in my Junk folder! Thanks for the explanation.

So here's another question: I see no theorem that says that the Fourier series for a function that is zero at zero and $x\sin(1/x)$ elsewhere between $-\pi$ and $\pi$ converges everywhere. Do we know whether it converges at zero? Eric Kvaalen (talk) 20:19, 8 September 2024 (UTC)

I don't know. $x\sin(1/x)$ is 1/2-Holder continuous, which is not quite enough for uniform convergence. For instance $x^{1+\epsilon }\sin(1/x)$ has a uniformly convergent Fourier series for $\epsilon >0$. Silly rabbit (talk) 23:28, 8 September 2024 (UTC)

@Silly rabbit: OK. Anyway, $x^{1+\epsilon }\sin(1/x)$ is of bounded variation in the interval, so the series converges by the Dirichlet–Jordan test. It seems to me that it would be difficult to show that $x\sin(1/x)?$ does not converge at zero. What about the derivative of $x\sin(1/x)?$ Its Fourier coefficients are well defined, but does it converge anywhere? Eric Kvaalen (talk) 10:13, 9 September 2024 (UTC)

Actually, the example of the Fourier series of $x\sin(1/x)$ does converge to 0 at $x=0$, because the function does satisfy a Holder condition (of exponent $0<\alpha <1/2$), by the Dini test. Silly rabbit (talk) 11:29, 9 September 2024 (UTC)

@Silly rabbit: Aha, you're right. (I think we can say $\alpha =1/2.$) However, that doesn't apply to its derivative. Eric Kvaalen (talk) 08:07, 10 September 2024 (UTC)

@Silly rabbit: Are there any continuously differentiable functions that are not already covered by being of bounded variation, so the Dirichlet-Jordan theorem applies? Eric Kvaalen (talk) 09:15, 16 September 2024 (UTC)